Wikipedia:Reference desk/Mathematics

Welcome to the mathematics section
of the Wikipedia reference desk.

skip to bottom

Select a section:

• Computing
• Entertainment
• Humanities
• Language
• Mathematics
• Science
• Miscellaneous
• Archives

Want a faster answer?

Main page: Help searching Wikipedia

   

How can I get my question answered?

• Select the section of the desk that best fits the general topic of your question (see the navigation column to the right).
• Post your question to only one section, providing a short header that gives the topic of your question.
• Type '~~~~' (that is, four tilde characters) at the end – this signs and dates your contribution so we know who wrote what and when.
• Don't post personal contact information – it will be removed. Any answers will be provided here.
• Please be as specific as possible, and include all relevant context – the usefulness of answers may depend on the context.
• Note:
  • We don't answer (and may remove) questions that require medical diagnosis or legal advice.
  • We don't answer requests for opinions, predictions or debate.
  • We don't do your homework for you, though we'll help you past the stuck point.
  • We don't conduct original research or provide a free source of ideas, but we'll help you find information you need.

Ready? Ask a new question!

How do I answer a question?

Main page: Wikipedia:Reference desk/Guidelines

• The best answers address the question directly, and back up facts with wikilinks and links to sources. Do not edit others' comments and do not give any medical or legal advice.

See also:

• Teahouse
• Help desk
• Village pump
• Help manual

May 13

Factoring Polynomials

Err, can someone explain to me how to factor polynomials? I can totally do x^2+bx+c but all of the sudden when I get ax^2+bx+c I just panic and resort to using the quadratic formula. I know there's a faster way to do it, but googling just gives me a whole bunch of stuff I don't understand. Can someone explain a method that's easy to understand?? Thanks! —Preceding unsigned comment added by 142.132.70.60 (talk) 19:21, 13 May 2011 (UTC)

You can factorize the expression by taking out the 'a' and get $ax^{2}+bx+c=a\left(x^{2}+{\frac {b}{a}}x+{\frac {c}{a}}\right)$. Then you can use your techniques on the term inside the big brackets; but don't forget to put the 'a' back in! HTH, Robinh (talk) 20:53, 13 May 2011 (UTC)

Personally, I find you have to start with the "x" term in each, for example (x _ _)(3x _ _), if you can fix the signs (if b>0, and c>0, then both signs are positive, if c>0 and b<0, both are negative) and then see if you can get it from there. If there are only a few possible combinations as factors of c then it's possible to write it out. (Also - a small hint, if you can get it in somewhere: if the brackets start with and even number of xs, then the second part of the bracket must be odd, else the whole thing would divide by two and from inspection is can't (unless you missed that originally)) Grandiose (me, talk, contribs) 21:19, 13 May 2011 (UTC)

$ax^{2}+bx+c=a\left(x-{\frac {-b+{\sqrt {b^{2}-4ac}}}{2a}}\right)\left(x-{\frac {-b-{\sqrt {b^{2}-4ac}}}{2a}}\right)$

The fast way is http://www.wolframalpha.com/input/?i=ax^2%2Bbx%2Bc

The easy-to-understand way is to check that $a\left(x-{\frac {-b+{\sqrt {b^{2}-4ac}}}{2a}}\right)\left(x-{\frac {-b-{\sqrt {b^{2}-4ac}}}{2a}}\right)$ is equal to $ax^{2}+bx+c$ Bo Jacoby (talk) 11:35, 15 May 2011 (UTC).

Latex

Can someone tell me how to draw the following in latex: a hollow square with a diagonal line going from bottom left corner to top right corner, and each line has an arrow head in the middle (not on the ends), so I want the lines to look like their pointing in some direction but I don't want the arrow head on the end points. I need the diagonal line's arrow head pointing 45 degrees clockwise from north. I also need the four vertices labelled as v. If you could post the code I'd appreciate it. Money is tight (talk) 22:32, 13 May 2011 (UTC)

By far the easiest thing to do is to draw the image in another program, and save it as a .png or .eps or whatever, and use \includegraphics{file}. Invrnc (talk) 23:05, 13 May 2011 (UTC)

You might try Inkscape to produce a .SVG. SVG is based on XML so if the drawing is simple and you're patient and willing to learn the syntax it's possible to create the file with a text editor.--RDBury (talk) 23:35, 13 May 2011 (UTC)

Alternatively, there are lots of packages for coding drawings in latex. PGF/TikZ is very good and has an extensive manual with lots of examples (linked from the article). In fact, the standard picture environment is probably enough for what you want to do - there is a nice Wikibooks article with lots of information about picture, that also discusses xy, tikz, xfig, and others. 81.98.38.48 (talk) 14:47, 14 May 2011 (UTC)

Also, in which direction are the other four arrowheads facing? – b_jonas 06:53, 15 May 2011 (UTC)

The left and right sides of the square are facing upwards, and the top and bottom sides of the square are facing right. Basically I'm trying to draw the triangulation of the torus. Money is tight (talk) 18:54, 15 May 2011 (UTC)

Here's an example (see thumbnail). See its description page for the EPS (encapsulated Postscript) formatted original that has infinite resolution and you can easily use in a LaTeX article.

I don't quite understand why you'd want an arrowhead on the diagonal if this is supposed to show the triangulation of the torus. I thought the arrowheads would mean identifying two edges on the drawing. Anyway, for this image I stick to your description so I've put an arrowhead to the diagonal as well, but feel free to modify it in any way you like (including changing the dimensions, font, etc). – b_jonas 21:50, 15 May 2011 (UTC)

Ah, I get it now. The four vertices are identified, so they should have the same label. I changed the image to have the same label (v) at each vertex. (Still don't get the arrow at the middle.) – b_jonas 21:57, 15 May 2011 (UTC)

Quoting part of OP's response from my talk page.

• I tried using the scripts you posted but I really don't know how to get them to work (I'm a total computer newbie). There's something more I need about the picture; I need the two vertical edges labelled a, the two horizontal edges labelled b, the diagonal edge labelled c, the top left triangle labelled U, and the bottom right triangle labelled V. Basically I'm trying to get it look like the first of three pictures on page 102 in Hatcher's book (http://www.math.cornell.edu/~hatcher/AT/ATch2.pdf).

I don't mind your replying on the talk page, and I will probably look at this a bit later, but allow me to copy the reply here as well so other helpful people reading here know about it. – b_jonas 08:37, 17 May 2011 (UTC)

File:Torus-triangulation-Hatcher.png

Figure extracted from Hatcher's book

You don't want to draw a figure, and you have found a good quality figure in an electronically typeset book that matches what you want. Why don't you ask permission to reuse that from the authors of that book? Here's the extracted image (see thumbnail on right). Again, the description page of that image contains the image in EPS (encapsulated Postscript) format, which you can easily include in a LaTeX document compiled with latex+dvips; or convert the image to pdf if you want to compile the LaTeX document with pdflatex. – b_jonas 10:16, 17 May 2011 (UTC)

Okay, wait a moment. I can't host the figure extracted from the book on the Wikipedia server, for it does not have a free license. So download the extracted figure in EPS format from here. The book comes with the following copyright notice:

• Copyright (c) 2002 by Cambridge University Press. Single paper or electronic copies for noncommercial personal use may be made without explicit permission from the author or publisher. All other rights reserved.

– b_jonas 16:25, 17 May 2011 (UTC)

Which are Martingales?

I know that there are two criteria for some $M_{t}$ to be an ${\mathcal {F}}_{t}$-Martingale.

Firstly $\mathbb {E} (|M_{t}|)<\infty$.

Secondly $\mathbb {E} (M_{t}|{\mathcal {F}}_{s})=M_{s}$.

Now $W_{t}$ is a Wiener process that generates a filtration ${\mathcal {F}}_{t}$.

I need to determine whether a couple of functions of $W_{t}$ are ${\mathcal {F}}_{t}$-Martingales or not.

The first one is $M_{t}=W_{t}^{2}$.

I'm assuming that this passes the first test. Because $Var(W_{t})=\mathbb {E} (W_{t}^{2})-(\mathbb {E} (W_{t}))^{2}=\mathbb {E} (W_{t}^{2})=\mathbb {E} (|W_{t}^{2}|)=t<\infty$

As for the second test, I'm guessing that it fails, since $\mathbb {E} (W_{t}^{2}|{\mathcal {F}}_{t})=\mathbb {E} (W_{t}^{2}-W_{s}^{2}+W_{s}^{2}|{\mathcal {F}}_{t})=\mathbb {E} (W_{t}^{2}-W_{s}^{2}|{\mathcal {F}}_{t})+W_{s}^{2}=(t-s)+M_{s}\neq M_{s}$

But the second one is much more difficult:

$M_{t}=tW_{t}-\int _{0}^{t}W_{s}ds$

I thought maybe $\mathbb {E} (|M_{t}|)=t\mathbb {E} (W_{t})-\int _{0}^{t}\mathbb {E} (W_{s})ds=0$

But I'm guessing there's something wrong about that. As for checking the second condition I really don't know where to begin. Thorstein90 (talk) 00:54, 14 May 2011 (UTC)

So, the first is not a martingale since it's expectation value is the variance of the Wiener process, which increases with time. The second is a martingale. The condition on the variance is straightforward:

$\operatorname {var} (M_{t})\leq E(M_{t}^{2})=t^{2}\operatorname {var} (W_{t})+2t\int _{0}^{t}E(W_{t}W_{s})ds+\int _{0}^{t}\int _{0}^{t}E(W_{s}W_{s'})\,ds\,ds'$

which you can estimate from above by the Cauchy-Schwarz inequality. For the martingale property, your calculation is basically right, except you need to condition on the initial information ${\mathcal {F}}_{s}$. Then

${\begin{aligned}E(M_{t}|{\mathcal {F}}_{s})&=tW_{s}-\int _{0}^{s}E(W_{s'}|{\mathcal {F}}_{s})\,ds'-\int _{s}^{t}E(W_{s'}|{\mathcal {F}}_{s})\,ds'\\&=tW_{s}-\int _{0}^{s}W_{s'}\,ds'-(t-s)W_{s}=M_{s}.\end{aligned}}$

-Sławomir Biały (talk) 12:20, 16 May 2011 (UTC)

Riemann Hurwitz formula and analytic maps on the Riemann Sphere

Hello everyone, I'm stuck on the following problem and have no idea how to get going on it, was hoping you could help.

Suppose $f:\mathbb {C} _{\infty }\to \mathbb {C} _{\infty }$ is an analytic map of degree 2: show that there exist Mobius transformations S, T such that $SfT:\mathbb {C} _{\infty }\to \mathbb {C} _{\infty }$ is the square map, $v\to v^{2}$.

Now I've just previous to this stated the Riemann Hurwitz formula which applies to f here, and I have also shown that the analytic isomorphisms of the Riemann sphere to itself are the non-constant mobius maps. So, I feel like I must be almost all the way there already, I just need help piecing this together to get the result: any thoughts anyone? Thank you! Totenines99 (talk) 23:59, 13 May 2011 (UTC)

Take a look at the complex analysis section of the ramification article. Then take a look at the statement section of the article on the Riemann–Hurwitz formula. Notice that you have been asked to consider the map ƒ(z) = z² which has a ramification of order two, and the Riemann sphere whose Euler characteristic is also two. Finally, the automorphism section of the Möbius transformation article tells us that the automorphisms of the Riemann sphere are the Möbius transformations. — Fly by Night (talk) 01:50, 14 May 2011 (UTC)

So I see that we can deduce from the formula that my f ramifies at exactly 2 points with multiplicity 2 in each case, right? So are we saying you can map these via T to any other 2 points on the Riemann sphere: do we want these to be 0 and infinity? Even if I have understood that far, I'm not quite following where you go from there, sorry.

I did have 1 further query while I'm waiting, if anyone can help me with it: I am trying to work out, with the Riemann surface R associated with complete analytic function $f(z)=(z^{2}-1)^{1/2}+(z^{2}-4)^{1/2}$ on $U=\mathbb {C} -\{\pm 1,\,\pm 2\}$ and regular covering map $\pi :R\to U$, what subgroup of the full symmetric group of $\pi ^{-1}(P)$ is obtained from all closed curves starting and ending at P (since these curves 'lift' to curves between preimages of P under the covering map, each gives a permutation of the preimages of the point P, via the starting/end point of the lifted curve). However, I'm having trouble figuring out how we can actually find out what permutations are possible via these curves: there are 4 preimages to every point under this map, right? (2 from the squaring map and two from the fact that the square root is a multi-valued function) So where do I go from here? I would really be very grateful for your help or advice on both of these problems. Totenines99 (talk) 14:24, 14 May 2011 (UTC)

No thoughts, anyone? :) Totenines99 (talk) 05:05, 17 May 2011 (UTC)

May 14

Statistics text

Hello. I'm looking for a good, rigourous, calculus-based introductory statistics text. It should start from the very basics (i.e., assuming you don't even know what a mean is [though I do], not basic as in starting from field theory because that is too basic :), kind of an equivalent to Spivak's calculus text. Does anyone have any recommendations? Thanks. 72.128.95.0 (talk) 15:37, 14 May 2011 (UTC)

I don't really know a lot of statistics textbooks, but you can try All of statistics. It starts at the beginning, but it's very condensed, it tries to cover a lot of technical ground without too much discussion. -- Meni Rosenfeld (talk) 08:28, 15 May 2011 (UTC)

De Groot. (Not "equivalent to Spivak's Calculus", but addressed to somewhat mathematically inclined readers.) Michael Hardy (talk) 22:02, 16 May 2011 (UTC)

Simple formula tom figure out kilobytes per second to number of hours?

Can anyone give me a simple formula to figure out how long it will take to download a file given a constant number of kilobytes per second and knowing the total size of the file in gigabytes? It would be easier for me to see the calculation I think with a real example so say I am downloading at a constant rate of 92 kilobytes per second and the file I am downloading is 13.4 gigabytes. I know I have to do something like multiplying the number of kbs by 60 to get to minutes and then 60 to get to hours and then dividing somehow into the number of gbs expressed in kbs but I get lost in the details. Thanks.--108.54.17.250 (talk) 15:12, 14 May 2011 (UTC)

You divide the file size by the download rate to get the number of seconds. Divide by 60 to get minutes, and by 60 again to get hours. In your example, you have 13.4 gigabytes = 13400000 kilobytes divided by 92 kilobytes/second to get 145652 seconds. Divide by 60 seconds/minute to get 2427 minutes. Divide by 60 minutes per hour for 40 hours. Detailed numbers are slightly different if you use base-2 gigabytes, instead of base 10, and if you don't just drop fractions. --Stephan Schulz (talk) 16:30, 14 May 2011 (UTC)

Great. Thank you. So its convert file size in gb → kb/kbs/60/60.--108.54.17.250 (talk) 16:42, 14 May 2011 (UTC)

To put it as simply as possible, divide the file size (in gigabytes) by the download rate (in kilobytes per second), and then multiply the result by 278 -- this will give you the time in hours. (Reason: 1 GB/hr = 278 kB/sec)Looie496 (talk) 17:00, 14 May 2011 (UTC)

Be careful mind, some transfer rates are given in kilobits per second (kbit/s, kb/s, or kbps) and there are 8 kilobits to the kilobyte. Grandiose (me, talk, contribs) 17:04, 14 May 2011 (UTC)

To further complicate things, often people write "gigabyte" when they really mean gibibyte. Also (though this may be irrelevant for the OP), sometimes the electrical bitrate is specified, and with 8b/10b encoding one data byte is 10 electrical bits. -- Meni Rosenfeld (talk) 08:21, 15 May 2011 (UTC)

Note that if you actually attempt to download 13.4 GB, your ISP is likely to stop you or slow the transfer rate down to a crawl. And, even if they don't, the download is likely to be interrupted anyway, so you need a way to restart an interrupted download. StuRat (talk) 04:17, 16 May 2011 (UTC)

Google: [1]. Sławomir Biały (talk) 12:09, 16 May 2011 (UTC)

May 15

Ellipse

The equation $ax^{2}+bxy+cy^{2}=1$ describes an ellipse, but it is not a standard ellipse because the ellipse's axes are not necessarily parallel to the x and y axes, i.e. it has been rotated. How do you read the angle of rotation from the equation? Widener (talk) 03:13, 15 May 2011 (UTC)

If the major axis forms an angle of $\theta$ with the x-axis, then θ minimizes the value of $ax^{2}+bxy+cy^{2}$ with the substitution $(x,y)=(\cos \theta ,\sin \theta )$. Substituting and differentiating gives $(c-a)\sin(2\theta )+b\cos(2\theta )=0$ which means $\theta ={\tfrac {1}{2}}\tan ^{-1}\left({\frac {b}{a-c}}\right)$. This gives the result up to a multiple of $\pi /2$. -- Meni Rosenfeld (talk) 08:17, 15 May 2011 (UTC)

Did this reply help you? -- Meni Rosenfeld (talk) 15:56, 16 May 2011 (UTC)

Yes. Why wouldn't it? You explained the answer to my question. Widener (talk) 11:48, 17 May 2011 (UTC)

It's common that even an answer as straightforward as this leaves an OP with something to be desired. I don't presume to divine what an OP would think upon seeing an answer of mine. I appreciate having explicit closure for the exchange. -- Meni Rosenfeld (talk) 10:01, 18 May 2011 (UTC)

Multiplication of Cardinals and Order Preservation.

See Multiplication#Properties. If a, b, and c are cardinal numbers, does the following still hold?

" Multiplication by a positive number preserves order: if a > 0, then if b > c then ab > ac. Multiplication by a negative number reverses order: if a < 0 and b > c then ab < ac."
Thanks in advance. voidnature 08:24, 15 May 2011 (UTC)

[ec]Are you referring to cardinal numbers? I don't know of a way to multiply a cardinal number with a negative number. For multiplication by a cardinal number a, this will hold if either $a<b$ or b is finite. -- Meni Rosenfeld (talk) 08:34, 15 May 2011 (UTC)

Yes, I am referring to cardinal numbers. My question is actually concentrated on this bit :"if a > 0, then if b > c then ab > ac". Thankyou. voidnature 08:36, 15 May 2011 (UTC)

Ok, so you need either $a<b$ or b finite. Otherwise, a counterexample is $a=\aleph _{0},b=\aleph _{0},c=1$. However, it will always be true that $ab\geq ac$. I think the axiom of choice might be needed for some of these results. -- Meni Rosenfeld (talk) 08:49, 15 May 2011 (UTC)

Oops, sorry, c has to be zero. So I don't need to worry about $a=\aleph _{0},b=\aleph _{0}$. So my question should be If a and b are cardinal numbers and c is 0, does "if a > 0, then if b > c then ab > ac" hold? Thankyou. voidnature 08:58, 15 May 2011 (UTC)

Every cardinal multiplied by 0 is 0, so you're basically asking, "if a>0 and b>0 are cardinal numbers, is ab>0?" The answer is yes, because the cartesian product of two nonempty sets is nonempty. -- Meni Rosenfeld (talk) 09:05, 15 May 2011 (UTC)

"the cartesian product of two nonempty sets is nonempty": can you give a proof please? voidnature 09:09, 15 May 2011 (UTC)

$a\in A,\ b\in B\Rightarrow (a,b)\in A\times B$. -- Meni Rosenfeld (talk) 10:17, 15 May 2011 (UTC)

Why is S5 a modal companion of CPC?

Why is S5 a modal companion of CPC? It seems like this should imply that S5 implies the translation of excluded middle, which seems to be $\Box p\lor \Box \lnot p$, which seems to say there are no contingent propositions - but surely S5 allows for contingent propositions? 88.104.173.35 (talk) 19:27, 15 May 2011 (UTC)

Plane partitioning algorithm

Resolved

Oh, this is on the tip of my tongue: I hate when that happens. Begins with "L", I think... I'm trying to remember the name of an algorithm which takes an array of points on a plane and partitions the plane such that each point is surrounded by a polygon - I think the margins of which fall equidistantly with another point (or is it some other definition of "influence"?). The resulting diagram looks like a honeycomb made by tipsy bees. What is that algorithm? (it's not a BSP or its ilk)-- Finlay McWalter ☻ Talk 22:22, 15 May 2011 (UTC)

Voronoi diagram perhaps?--RDBury (talk) 23:39, 15 May 2011 (UTC)

Yes, that's it (and no L in sight)! Thanks. -- Finlay McWalter ☻ Talk 23:45, 15 May 2011 (UTC)

“L, I KNOW it begins with L!” – b_jonas 18:46, 16 May 2011 (UTC)

May 16

Black Scholes Transformation

This is a basic calculus question. I'm reading this book and on page 102 there is the Black Scholes equation

${\frac {\partial V}{\partial t}}+{\frac {\sigma ^{2}(t)}{2}}S^{2}{\frac {\partial ^{2}V}{\partial S^{2}}}+r(t)S{\frac {\partial V}{\partial S}}-r(t)V=0$

and the following substitutions are made

${\bar {S}}=Se^{\alpha (t)}$

${\bar {V}}=Ve^{\beta (t)}$

${\bar {t}}=\gamma (t)$

to turn the original equation into the following:

${\frac {d\gamma }{dt}}{\frac {\partial {\bar {V}}}{\partial {\bar {t}}}}+{\frac {\sigma ^{2}(t)}{2}}{\bar {S}}^{2}{\frac {\partial ^{2}{\bar {V}}}{\partial {\bar {S}}^{2}}}+(r(t)+{\frac {d\alpha }{dt}}){\bar {S}}{\frac {\partial {\bar {V}}}{\partial {\bar {S}}}}-(r(t)+{\frac {d\beta }{dt}}){\bar {V}}=0$

What's going on here exactly? I can't put my finger on where the terms in the transformations are coming from, I know it must be some sort of application of the chain rule but I don't see what exactly. —Preceding unsigned comment added by 130.102.158.15 (talk) 04:30, 16 May 2011 (UTC)

The formula you've listed is the partial differential equation version of the Black-Scholes formula. I believe the substitutions establish the boundary conditions for the solution. You can find more about this on page 440, here: http://faculty.atu.edu/mfinan/actuarieshall/DFEM.pdf. —Preceding unsigned comment added by 12.186.80.1 (talk) 20:16, 16 May 2011 (UTC)

Your best off working backwards. Go through the second equation doing all the differentiation and you should see that you get back to the first equation. (Assuming it's correct, I haven't actually checked it.) --Tango (talk) 20:20, 16 May 2011 (UTC)

Area enclosed by ellipse equation

How do you find the area enclosed by the ellipse $ax^{2}+bxy+cy^{2}=1$ assuming the coefficients are chosen such that the equation yields an ellipse (rather than a hyperbola). Widener (talk) 06:40, 16 May 2011 (UTC)

Per ellipse, the area is ${\frac {2\pi }{\sqrt {4ac-b^{2}}}}$. Dragons flight (talk) 06:49, 16 May 2011 (UTC)

Fancy that! Is there a derivation or proof? Widener (talk) 06:59, 16 May 2011 (UTC)

The proof is about two lines if you use the fact that the discriminant is an invariant on binary quadratic forms. I don't think they cover that much in undergraduate curricula nowadays though.--RDBury (talk) 12:02, 16 May 2011 (UTC)

Generating sequences based on probabilities

I want to create a set of fixed-length sequences based on an a priori set of probabilities describing the distribution of symbols in the resulting set of sequences. The case where the probability of each position is independent is simple enough (simply pick each symbol randomly according to probability), but I'm at a bit of a loss as how to approach the problem if the input probabilities also specify the joint probability of multiple positions.

Perhaps a toy example will make things clearer. Take a three position binary sequence A,B,C. The input to the problem is probabilities for a symbol at each position, e.g p(A=1) = 0.6; p(B=1) = 0.5; p(C=1) = 0.55. Given those probabilities, I want to create a set of output sequences that, in the limit, would satisfy those probabilities. While simple enough for independent positions, I'm also interested in cases where joint probabilities are also specified, e.g. p(A=1,B=1) = 0.3; p(B=1,C=0) = 0.2; p(A=1,C=1) = 0.25. I'm not sure how to generate the sequence when there's not just a single probability to satisfy per position, but a web of interlocking probabilities. - As I said, that's just a toy example for explanatory purposes. The cases that I'm looking into are more complicated, with potentially dozens of positions and dozens of symbols at each position, so enumerative approaches aren't going to work. The joint probabilities would be limited to pairs of positions, though. (But approaches that are extensible to triples or quartets of positions would be all the better.) Any suggestions or pointers? Thanks. -- 140.142.20.229 (talk) 17:32, 16 May 2011 (UTC)

This is not solvable on the basis of the information provided. Without some extra information such as independence or second-order-independence, the joint probabilities do not determine the global probabilities, and you need to know the global probabilities in order to generate exemplars. (Bayesian techniques sometimes make use of joint probabilities of this sort, but they are not guaranteed to give the correct answer.) Looie496 (talk) 17:41, 16 May 2011 (UTC)

I'm not familiar with the concept of "second-order-independence", so am I right in assuming that it's the three event version of two-event independence (e.g. p(X,Y) = p(X)p(Y))? If so, yes, I should have stated that that can be assumed. Basically, the given probabilities are the only constraints on the system. Having the other degrees of freedom being as unbiased as possible is preferred, although if biasing them slightly is a way to make the problem more tractable, I'm open to it. As I mentioned, the problem is somewhat trivial if you can assume (pairwise) independence, but one of my issues is that I'm not sure how the three-plus event version of independence would behave (where there's nothing special about the probability distribution of triples of events that isn't captured by the probabilities of the individual events and pairs of events; like with pairwise p(X,Y) = p(X)p(Y) independence there's nothing about the probability of the pair of events that isn't captured by the probabilities of the single events), or even if it would simplify the problem. -- 140.142.20.229 (talk) 20:40, 16 May 2011 (UTC)

I would do this by first having a validity check - if any position has more than one symbol, the sequence is not valid. Then, calculate each probability. Say you set A=1/B=1 in the first probability and then you set B=0/C=1 in the following probability. This is not a valid sequence because B cannot be 1 and 0 at the same time. Quit and try again. Set that in a loop to try it over and over and over until a valid sequence appears. I just used this tactic because I lost my key for a piece of software. I knew the range and probabilities of most of the elements of the key, so I wrote a program to randomly generate keys until one of them passed the software's key check. In your cause, the validity check would be the same as the key check I used. -- kainaw™ 17:59, 16 May 2011 (UTC)

If I'm understanding you correctly, you're suggesting to select A,B, and C based on their individual probabilities, and then go through the list of pair probabilities, and then do random number draws and checks for each one? For example, We randomly pick ABC to be 110. We then look at the AB pair, draw a random number, say 0.25, find it passes the p(A=1,B=1) = 0.3 check, and move on to the BC pair. We pull another random number, say 0.65, and find it fails the p(B=1,C=0) = 0.2 check. We then throw out the selection, and regenerate an ABC sequence from scratch. - My concern with that is with a large number of pair probabilities, the procedure would be highly inefficient as the repeated tests mean most generated sequences would randomly fail at least one of the pair probability tests. -- 140.142.20.229 (talk) 20:51, 16 May 2011 (UTC)

And I'm not sure this will converge to the correct result. -- Meni Rosenfeld (talk) 09:46, 17 May 2011 (UTC)

It does because you did it below. You just described it better. You produce a lot of candidates and you pick the ones that maintain the expected distribution (I called the function that checked the expected distribution a "validity check"). I wrote it in a tiny script that produced 1,000,000 candidates. I got 420,028 sets to pass the validity check. My resulting percentages are 0.600, 0.502, 0.551, 0.245, 0.300, 0.211. I noticed that it is always closer to the expected percentages in the single element checks than the two-element checks. Also, it never gets A=1/C=1 down to 0.200. It is always high (once it was 0.232). -- kainaw™ 16:39, 17 May 2011 (UTC)

Ok, I get your method now. I think the OP's interpretation of it is wrong. It is very different from my method. I'm still not totally convinced that it works. -- Meni Rosenfeld (talk) 19:06, 17 May 2011 (UTC)

I see a possible mistake in the implementation in which the questioner stated. Simply, it states that there is a 30% chance A=1 and B=1. That means that there is a 70% chance that A=0 or B=0. You need to check both of those. Otherwise, it is not functionally the same as using a counter/weight for each of the probabilities. By using a counter/weight, you checked both sides of the probability. -- kainaw™ 19:14, 17 May 2011 (UTC)

I think the following should work. For each constraint, keep a running total of how many times its condition held in the items generated so far. When you need to generate a new item, generate one independently randomly based on the single-position constraints. Calculate the sum-of-squared-differences between the required probabilities (including single-position) and the empirical probabilities, for the current set and what your set would become if you added the new sample. Find out how much the new item improves your SSE, and either add it or reject it with a probability that depends on the improvement (maybe a linear function with cutoffs at 0 and 1, and no improvement -> 0.5 probability). -- Meni Rosenfeld (talk) 09:46, 17 May 2011 (UTC)

Actually, the probability for 0 improvement should be less than 0.5. Hold on... -- Meni Rosenfeld (talk) 10:05, 17 May 2011 (UTC)

Ok, new plan, even better than the last one. Keep a weight parameter for each condition (eg "A=1, B=1"), initialized to 0. When you generate a candidate item, sum up the weights for the conditions it matches. Include it with probability equal to the logistic function of the sum. If you do include it, add to each weight some number times the difference between the target probability for the condition (for "A=1, B=1" this is 0.3) and the indicator function of whether the item matches the condition.

I've simulated this on your toy example, and after roughly 100,000 candidates, I have a set with 83548 items and the respective empirical probabilities are {0.590822, 0.498731, 0.549911, 0.301156, 0.201393, 0.259168}, compared to the target {0.6, 0.5, 0.55, 0.3, 0.2, 0.25}.

You can also in the end discard all items generated before the weights stabilized. -- Meni Rosenfeld (talk) 10:27, 17 May 2011 (UTC)

Here's the Mathematica code for my simulation:

conditions = {{1, _, _}, {_, 1, _}, {_, _, 1}, {1, 1, _}, {_, 1, 0}, {1, _, 1}};
target = {0.6, 0.5, 0.55, 0.3, 0.2, 0.25};
lgs[x_] := 1/(1 + Exp[-x]);
emp[X_] := N[Map[Count[X, #] &, conditions]/Length[X]];
matches[x_] := Map[Count[{x}, #] &, conditions];
set = {};
weights = Table[0, {Length[conditions]}];
Do[If[lgs[weights.matches[#]] > RandomReal[], AppendTo[set, #];
 weights += 0.1 (target - matches[#])] &[RandomInteger[1, 3]], {100000}]

-- Meni Rosenfeld (talk) 11:43, 17 May 2011 (UTC)

Thanks, that seems to work decently. I was a little concerned that it might be slow for large number of low-probability conditions (as you're likely to get if you have dozens of symbols at dozens of positions all extensively connected), but it seems like the acceptance rate stays above 50%, even for the small probability cases. Is there a name for this method (e.g. is it a modification of some standard technique)? I'm wondering how best to refer to it when discussing it with others. -- 140.142.20.229 (talk) 21:00, 17 May 2011 (UTC)

I don't know a name for it, but I was inspired by logistic regression when thinking about it.

Actually, I have doubts now that it is guaranteed to work. If the weights stabilize around some finite numbers, then the generated accepted items will satisfy the conditions. But the weights will likely not stabilize, so the result could be a complex interplay of the way it diverges to infinity.

It might help somewhat if you allow a constant parameter added to the sum. I'm not sure how it should be calibrated.

It's an interesting problem, but I don't have the time to continue analyzing it. Play around with it a little and see what you get. -- Meni Rosenfeld (talk) 08:17, 18 May 2011 (UTC)

It is generally called rejection sampling. Using weights is one method of rejection sampling. The drawback is that the results are not accepted to be random because the generation of a new sequence is based on the previous sequences that were generated. Sort of like flipping a coin 5 times and getting 3 heads so you purposely force it to go tails to get a 3-3 distribution. -- kainaw™ 12:28, 18 May 2011 (UTC)

Let me suggest a useful notation for the conditions. Let the values be called 1 and 2 (rather than 1 and 0), and let the variables ABC be indicated by the digit position. Thus the conditions (A=1) (B=1) (C=1) (A=1,B=1) (B=1,C=0) (A=1,C=1) are called 100 010 001 110 012 101. Of course you have p(000)=1, so you can compute the probabilities of 200 020 002 120 210 220 011 102 and 101, but you do not know the probabilities for 021 and 022 individually, nor the probabilities for 111 112 121 122 211 212 221 and 222. Bo Jacoby (talk) 13:22, 18 May 2011 (UTC).

Normal subgroups

I'm trying to study for an exam, and this is one of the review questions listed on the textbook author's website:

Q: If N is a normal subgroup and |aN| = n, then a^n = e. 
A: false
Remark: a^n belongs to N.

I can't see for the life of me why a^n should belong to N. Isn't |aN|=n just |N|, regardless of what a is, and regardless of whether N is normal? By substitution, that would seem to imply that a^(|N|) belongs to N, for any element a and for any subgroup N, which can't possibly be right... 71.176.168.114 (talk) 19:35, 16 May 2011 (UTC)

|aN| doesn't necessarily equal |N|. Consider N=Z_2 (ie. integers mod 2, so it has order 2) and a=2. aN is then the trivial group, so has order 1. --Tango (talk) 20:27, 16 May 2011 (UTC)

OK, you're right, but I still don't see the leap to "a^n belongs to N." Can you explain that or give me a hint? 71.176.168.114 (talk) 20:32, 16 May 2011 (UTC)

I think that |aN| = n is supposed to mean that the order of the element aN in the quotient group G/N is n. Matthew Auger (talk) 21:09, 16 May 2011 (UTC)

Thank you 71.176.168.114 (talk) 21:15, 16 May 2011 (UTC)

Surely |aN|=n simply means that the group aN has n elements. It may follow from that that [a] in G/N has order n (from which it very quickly follows that a^n is in N), but it still needs to be proven. --Tango (talk) 23:04, 16 May 2011 (UTC)

aN would only be a group if a∈N. The notation is confusing but I think the only interpretation that makes sense mathematically is |aN| means the order of aN in G/N. As an example, take G=<a>=cyclic of order 4, N=<a²>. Then G/N=<aN> is cyclic of order two but a² is not e.--RDBury (talk) 01:33, 17 May 2011 (UTC)

Is there a word for a (position, orientation) pair?

It is something I use so commonly in geometry I'm amazed that after Googling I have not found a good word for it. Together the two describe 6 degrees of freedom of an object in relation to a defined axis system, typically

X, Y, Z, yaw, pitch and roll

There are many different ways of representing both the position and orientation parts. Any suggestions, no matter how informal, are welcome as long as it makes intuitive sense - I'm trying to choose a good type name in a program I'm writing.

196.215.115.184 (talk) 20:21, 16 May 2011 (UTC) Eon

Congruence, isometry, isometric map, movement? – b_jonas 20:29, 16 May 2011 (UTC)

They are examples of generalized coordinates. I don't know a more specific term. --Tango (talk) 20:32, 16 May 2011 (UTC)

Thanks, I think configuration is the best general word I could find so far after following your links. Actually, I thought posientation is a cool word to describe that. It turns out I'm not the first person on the internet to have thought of that! 196.215.115.184 (talk) 20:59, 16 May 2011 (UTC) Eon

What about pose? It's a term that comes up a lot in engineering when speaking of such matters.--Leon (talk) 21:59, 16 May 2011 (UTC)

Brilliant! That's the best word for it. And this is a computer vision application so I'm convinced. Thanks so much for your input. 196.211.175.2 (talk) 08:24, 17 May 2011 (UTC) Eon

Linear map from unit circle to ellipse

What is the linear transformation that maps the relation $x^{2}+y^{2}=1$ onto $ax^{2}+bxy+cy^{2}=1$ Widener (talk) 22:30, 16 May 2011 (UTC)

Try completing the square. That should give you the equation in a form where it is easier to see the necessary transformation. --Tango (talk) 23:00, 16 May 2011 (UTC)

Thanks. Widener (talk) 11:57, 17 May 2011 (UTC)

By the way, your question implies that the answer is unique, but it's not. Any 2×2 orthogonal matrix $Q$ maps the unit circle to itself, so if you have a linear transformation $T$ that maps the unit circle to an ellipse, then $TQ$ is another solution. 98.248.42.252 (talk) 16:21, 17 May 2011 (UTC)

May 17

Life density

In Conway's Game of Life, what's the maximum (or maximum known) density for stable (or oscillating) periodic patterns? For oscillators, I'll accept either the maximum or the minimum density in the cycle as "the density".

Mere curiosity here, no higher purpose. —Tamfang (talk) 03:06, 17 May 2011 (UTC)

This is mentioned in the Life Lexicon and its been proved the max density is a half. --Salix (talk): 08:50, 17 May 2011 (UTC)

Some nice examples can be found at LifeWiki: Agar.--Salix (talk): 08:55, 17 May 2011 (UTC)

EXPAND MULTINOMIAL EXPRESSION

HOW DO I EXPAND THESE BINOMIAL EXPRESSION - (1+2X-X^2)^5 —Preceding unsigned comment added by 202.63.228.211 (talk) 03:42, 17 May 2011 (UTC)

1+10x+35x^2+40x^3-30x^4-68x^5+30x^6+40x^7-35x^8+10x^9-x^10 Wikinv (talk) 04:56, 17 May 2011 (UTC)

Giving the worked-out expression doesn't answer the question "how do I..", and more generally doesn't help the questioner solve similar (homework?) problems in the future. AndrewWTaylor (talk) 11:47, 17 May 2011 (UTC)

LIKE SO.

(1+2X-X^2)^5

(   1+2X-X^2   )^5

(     1+2X-X^2     )^5

(       1+2X-X^2       )^5

(         1+2X-X^2         )^5

(           1+2X-X^2           )^5

(             1+2X-X^2             )^5 —Preceding unsigned comment added by 72.179.51.84 (talk) 13:07, 17 May 2011 (UTC)

http://www.wolframalpha.com/input/?i=(1%2B2X-X^2)^5 Bo Jacoby (talk) 14:53, 17 May 2011 (UTC).

Polynomial expansion should be a good place to start. You could work up the powers finding $(1+2X-X^{2})^{2}$ first then multiplying that by $1+2X-X^{2}$ etc. Or let $y=2x-x^{2}$ and use the binomial theorem on $(1+y)^{5}$ or indeed jump straight to the Multinomial theorem.--Salix (talk): 21:50, 17 May 2011 (UTC)

Making fake fields with known covariance

I have a problem where I would like to simulate instances of a random data field given known statistical properties.

Specifically, $y({\vec {x}},t)$ is a random variable with a time-averaged expectation of zero at all locations, and

$\mathrm {Cov} (y({\vec {x}}_{1},t),y({\vec {x}}_{2},t))=C({\vec {x}}_{1},{\vec {x}}_{2})$

$\mathrm {Cov} (y({\vec {x}},t),y({\vec {x}},t-t_{0}))=T({\vec {x}},t_{0})$

Where $C({\vec {x}}_{1},{\vec {x}}_{2})$ and $T({\vec {x}},t_{0})$ can be assumed to be known for all locations and times.

How would one go about creating simulated fields for $y({\vec {x}},t)$ (over a set of grid locations) that obeyed the specified properties? I suspect (but am not entirely sure) that specifying C and T is not sufficient to completely constrain the statistical properties of the underlying probability distribution. If so, what other assumptions / approximations might be necessary. Dragons flight (talk) 10:17, 17 May 2011 (UTC)

You might find Correlation and dependence#Correlation and linearity useful or perhaps Normally distributed and uncorrelated does not imply independent. Dmcq (talk) 12:42, 17 May 2011 (UTC)

One way to do it is to use a Gaussian process. HTH, Robinh (talk) 20:44, 17 May 2011 (UTC)

Mobius transformations and concentric circles

Hello everyone,

In a course I'm taking, the lecturer has stated that if we have 2 circles which do not intersect at any point in the complex plane (lines also considered as circles), then we can take a mobius map mapping these 2 circles to concentric circles (e.g. |z|=1, |z|=R). Now it is obvious that if we can map to 2 concentric circles then our original 2 circles cannot touch, but it is not obvious to me why there necessarily exists such a transformation. I am aware that there is always a mobius map between any 2 circles on the Riemann sphere, but why is there necessarily one which will also make our 2 given circles concentric? (I mean 'concentric' in the sense that when they are seen in the complex plane, they are concentric around the origin.) Could anyone give me a very brief proof or explanation?

Thanks very much 131.111.185.74 (talk) 19:46, 17 May 2011 (UTC)

I think any pair of circles must be concentric on the Riemann sphere. You could rotate the sphere so that infinity lies in the centre of one circle, projecting to the plane would ensure that they images were concentric.--Salix (talk): 06:11, 18 May 2011 (UTC)

Sphere rotations only have 3 degrees of freedom. But you'd need 3 to map the first circle to the unit circle and 2 more to map move the center of the second to the origin. I'm pretty sure it can be done but group of Möbius transformations has 6 degrees of freedom and you need all but one of them; one degree of freedom use used up by a nontrivial subgroup that leaves both circles invariant. You know any circle can be mapped to the unit circle, and inversion swaps the inside with the outside, so you can assume the first circle is already the unit circle and the second circle is inside it. The Möbius transformations that preserve the unit circle are basically (though not quite) the isometries of the hyperbolic plane so it's really a matter of showing that there is an isometry that takes any circle to a circle with center at the origin. In other words what you would actually be showing is that circles on the plane are actually hyperbolic circles when considered as curves in the hyperbolic plane. See Poincaré disk model but the article does not mention this.--RDBury (talk) 08:25, 18 May 2011 (UTC)

The standard way to prove this is to put the circles into a normal form. I would map the first circle to the x axis such that the common diameter of the two circles goes to the y axis, and then use the remaining real scaling degree of freedom. Sławomir Biały (talk) 13:03, 18 May 2011 (UTC)

Probability Question

Is it possible to calculate the chance of getting a certain empirical probability relative to a given theoretical probability? For example, if I have a theoretical probability of about 25.5% but I get the result only 8.3% of the time. Do you understand what I am asking at all? Ryan Vesey (talk) —Preceding undated comment added 22:06, 17 May 2011 (UTC).

You're not being very clear, but let me guess: You took a random sample—in effect performed some Bernoulli trials—from a population where success occurs 25.5% of the time, and in your sample you got a success 8.3% of the time, i.e. about 1 out of 12. Certainly the probability of getting exactly one success, or at most one success, in 12 independent trials, with probability 25.5% of success on each trial, can be computed easily. It's a routine exercise. See binomial distribution. On the other hand, you could have had 2 successes, or at most 2 successes, in 24 trials, in which case the probability would be different.

But you shouldn't be making people guess what you mean. Michael Hardy (talk) 02:13, 18 May 2011 (UTC)

Well, I think the meaning of the question is reasonably clear, even though the terminology is incorrect -- Ryan is asking about the chance of getting a certain observed frequency given a certain probability distribution. It's a little harsh to demand that people who are confused about something be able to state their confusion in correct terminology. (The answer that Michael gave is the correct one, though: the probability is given by a binomial distribution.) Looie496 (talk) 02:34, 18 May 2011 (UTC)

May 18

Rubik's Cube Group Theory!

If you take a sequence of turns on a Rubik's cube and repeat them $n$ times without changing orientation of the cube between applications, then you are effectively computing $g^{n}$, where $g$ is the element of the Rubik's Cube group representing that sequence. It is a fact of group theory, then, that you are guaranteed to return to solved position eventually by repeating the sequence.

Question: is there a nontrivial example of a sequence of turns which will reach a solved position in the middle of one of the applications of the sequence of turns? More formally, if $g=t_{1}\circ \cdots \circ t_{i}$, where $t_{i}$ are single turns of a face, then we know $g^{n}=e$ for some $n$, but is there the possibility that $g^{j}\circ t_{1}\cdots \circ t_{k}=e$ where $1<k<i$?

Examples of trivial cases would be if the sequence of turns is, say, just 3 90° rotations of the same face clockwise; here it takes 4 applications to get back to identity, but it really only takes 1 application and then the turn of the next application. Another example would be if the sequence of turns itself returns to identity before reaching something else. I haven't made a precise formulation of what 'trivial' means here, and reserve the right to expand on what should be considered trivial if shown some further counterexamples. --72.179.51.84 (talk) 02:47, 18 May 2011 (UTC)

One possible first pass at making my first trivial case more precise would be to require that the sequence of turns not itself be a repetition of a sub-sequence more than one time. --72.179.51.84 (talk) 02:54, 18 May 2011 (UTC)

You are guaranteed to return to the original position eventually by repeating the sequence. You are not guaranteed to return to the solved position eventually by repeating the sequence. Bo Jacoby (talk) 06:03, 18 May 2011 (UTC).

Above is right. If you take a really jumbled up cube and g is a single face turn, then doing gⁿ will never give a solved cube. It will just cycle between 4 different cube states. Staecker (talk) 11:33, 18 May 2011 (UTC)

I think the OP meant that he starts with the solved position... -- Meni Rosenfeld (talk) 12:26, 18 May 2011 (UTC)

I think Meni Rosenfeld is right, but who am I to judge what the OP meant? --72.179.51.84 (talk) 12:33, 18 May 2011 (UTC)

If you do not usurp yourself, you are. -- Meni Rosenfeld (talk) 14:06, 18 May 2011 (UTC)

Pick any group element $g$. Let n be its order. Pick any integer $j$ where $0<j<n$. Let $t_{1}\cdots \circ t_{k}$ be a solution to $g^{-j}$ found by Optimal_solutions_for_Rubik's_Cube (so $k\leq 20$). Let $t_{k+1}\cdots \circ t_{i}$ be a solution to $g^{j+1}$ found by the optimal algorithm. Then $g=t_{1}\circ \cdots \circ t_{i}$ satisfies your conditions where $i\leq 40$. The example should be non-trivial for most choices of $g$ and $j$. 98.248.42.252 (talk) 14:57, 18 May 2011 (UTC)

Thanks, that makes sense. --72.179.51.84 (talk) 13:18, 19 May 2011 (UTC)

sum

∑Failed to parse (syntax error): {\displaystyle \sum_{v=∈V}^{dv} i}

Arfle barfle gloop? Dmcq (talk) 11:50, 18 May 2011 (UTC)

financial accounting 1

Mr mubaiwa a sole trader drew up his trial balance on 31/02/08. As it would not balance, a suspense account was opened and the difference debited to it. The trial balance is shown below:

                             $            $
Premises                   20 000
Buildings                  30 000
Motor vehicles             10 000
Purchases                  20 000
Stock opening               1 000
Debtors                     6 700
Cash at bank                6 140  
Cash in hand                  760  
Discount allowed              900
General expenses            1 500 
Rent                        1 400
Wages                       1 208
Capital                                 61 000
Sales                                   35 000
Discount received                        4 000
Suspense account              392             
                             ----
                          100 000      100 000
                             ----         ----

Closing stock as at 31/01/08 is $7 000: Subsequently the following errors were discovered

a)The wages had been overcast by $120
b)Sales had been overcast by $100
c)The purchase of machinery was posted to purchases account $2 000
d)Discount received $143 had been posted to the account as $134
e)Discount allowed $200 had been posted to the wrong side of the account
f)Bank charges $21 had been posted to the ledger.
g)Wages of $3 000 had not been entered into the books.

Required:

a)Prepare journal entries to correct the above errors.  (8 marks)
b)Write up the suspense account, clearing the balance in the account and correcting the errors.  (6 marks)  

—Preceding unsigned comment added by 209.88.92.29 (talk) 09:08, 18 May 2011 (UTC)

We're not just going to do your homework for you. First you have to try it, then show us what you did and ask for help where needed. StuRat (talk) 09:35, 18 May 2011 (UTC)

an "elementary calculation"

Hi. I'm reading a paper, and stuck on the following:

If $x$, $y$ are coprime integers and $n>1$, an elementary calculation shows that:

$\left({\frac {x^{n}\pm y^{n}}{x\pm y}},x\pm y\right)\mid n$.

One writes $x=(x\pm y)\mp y$ and then expands.

How does that work? I tried writing down what it says and raising it to the power $n$, but that doesn't seem to help. Any ideas? Thanks in advance if anyone can shed some light here. -GTBacchus^(talk) 16:15, 18 May 2011 (UTC)

Actually, I've got it worked out now, except in the case where $n$ is even and we're looking at ${\frac {x^{n}+y^{n}}{x+y}}$. The ${\frac {x^{n}-y^{n}}{x-y}}$ case was easy, and so was the plus case when $n$ was odd. -GTBacchus^(talk) 16:33, 18 May 2011 (UTC)

help please

Simplifying a polynomial fraction

How do I turn (3-3x)/[(3x-2)(x-1)] into -3/(3x-2)? When I distribute the denominator I get 3x^2-5x+2, so I guess that's wrong since I somehow need 3x-2 for the denominator. —Preceding unsigned comment added by 142.132.6.81 (talk) 17:38, 18 May 2011 (UTC)

$3-3x=3(1-x)=-3(x-1)$ Readro (talk) 17:42, 18 May 2011 (UTC)

(ec) I added a more useful (sub)title. Ok, let me rewrite it first:

  (3-3x)
-------------
 (3x-2)(x-1) 

Now take a 3 out of the numerator:

  3(1-x)
-------------
 (3x-2)(x-1) 

Make that a negative 3:

  -3(-1+x)
-------------
 (3x-2)(x-1) 

Reorder the portion of the numerator in parens:

  -3(x-1)
-------------
 (3x-2)(x-1) 

The (x-1) cancels:

   -3
--------
 (3x-2) 

We're done. StuRat (talk) 17:48, 18 May 2011 (UTC)

I was trying so hard to change the denominator when it was just a thing with the numerator all along? Thanks Readro and thanks so much for taking it all apart StuRat!142.132.6.81 —Preceding unsigned comment added by 142.132.6.81 (talk) 19:02, 18 May 2011 (UTC)

Diophantine equations

Hello. WHat are some methods to solve simple Diophantine equations? I saw your article and I don't think I understood it very well, to a large part I suspect because it was being more rigorous than I need, and accounting for possibilities or special cases that I personally am unlikely to encounter. I would like a brief walk-through or a link to a site with a brief walk through, if possible. Thanks. 72.128.95.0 (talk) 23:55, 18 May 2011 (UTC)

May 19

Independence of Wiener process increments (expectation of product of random normal variables)

If $z_{j}=W_{t_{j}}-W_{t_{j-1}}$ and $z_{k}=W_{t_{k}}-W_{t_{k-1}}$ are two non-overlapping Wiener process increments with distribution ${\mathcal {N}}(0,t_{j}-t_{j-1})$ and ${\mathcal {N}}(0,t_{k}-t_{k-1})$ respectively, how do I prove their independence?

I know that they will be independent if they have zero covariance, and that $\operatorname {Cov} (z_{j},z_{k})=\operatorname {E} {\big [}z_{j}z_{k}{\big ]}-\operatorname {E} [z_{j}]\cdot \operatorname {E} [z_{k}].$

And I know that $\operatorname {E} [z_{j}]=\operatorname {E} [z_{k}]=0$

So I guess that all I have to show is that $\operatorname {E} {\big [}z_{j}z_{k}{\big ]}=0$

Now according to this link the product of two normal distributions has some sort of weird distribution with a Bessel function and a Dirac delta function in it, and I can't see how to take the expectation of this.

I would appreciate any advice about how to figure this out. —Preceding unsigned comment added by Thorstein90 (talk • contribs) 00:06, 19 May 2011 (UTC)

It is certainly not true that any two random variables with zero covariance are independent. For example, suppose X = −1, 0, or 1 with equal probabilities and Y = X². Then the cov(X,Y) = 0 but X and Y are obviously not independent.

But as for proving that increments of a Wiener process on non-overlapping intervals are independent: that's part of the definition. If the increments are not independent, then it's not a Wiener process. Michael Hardy (talk) 03:01, 19 May 2011 (UTC)

Sorry, I better restate this. My question does not define them as Weiner process increments because that is what we are trying to prove. Instead we are given the condition

$\mathbb {P} (W_{t_{1}}\in A_{1},\dots ,W_{t_{n}}\in A_{n})=\int _{A_{1}}\dots \int _{A_{n}}p(x_{1},0,t_{1})p(x_{2},x_{1},t_{2}-t_{1})\dots p(x_{n},x_{n-1},t_{n}-t_{n-1})dx_{1}\dots dx_{n}$

where $p$ is the Gaussian transition probability, and we already know that the increment $W_{t}-W_{s}$ is normally distributed with mean 0 and variance $t-s$. From this we need to show that the increments are independent.

I'm not sure if I should be attempting to just show that

$\mathbb {P} ((W_{j}-W_{j-1})\in A_{j},(W_{k}-W_{k-1})\in A_{k})=\mathbb {P} (W_{j}-W_{j-1})\mathbb {P} (W_{k}-W_{k-1})$

or using the covariance condition instead. —Preceding unsigned comment added by Thorstein90 (talk • contribs) 05:27, 19 May 2011 (UTC)

The formula involving the Bessel function that you linked to applies only if it is already known that the two random variables are independent. Michael Hardy (talk) 03:06, 19 May 2011 (UTC)